A figure shows a square made from four right-angled triangles that all have the same dimensions. Each of the four right angled triangles have a height labeled a, a base labeled b, and a hypotenuse labeled c. Sides a and b are positioned so that the right angle creates the four outer corners of the outer square. Each vertex of the inner square divides each side of the outer square in two unequal parts labeled a and b, where a is the shorter part and b is the longer part. Each side of the inner square, labeled c, becomes the hypotenuse of the four right-angled triangles formed at the corners of the outer square. The four right-angled triangles are shaded.
Betty is writing a proof of the Pythagorean Theorem using the diagram shown. In the diagram, there is a large square. A smaller square is drawn inside the large square, dividing the large square into a smaller square and 4 small triangles.
Step 1: First, I know that the area of a triangle is equal to bh2
. The area of each small triangle would be ab2
. Since there are 4 triangles in the entire square, the area of all 4 triangles together would be equal to 2ab
. The area of the small square in the middle can be represented by c2
since the sides each have a length of c
. Therefore, the area of the entire large square can be written as 2ab+c2
.
Step 2: Another way to think about the area of the entire large square is by looking at its side lengths. The length of each side can be written as a+b
, so the area would be (a+b)×(a+b)
.
Step 3: Since we've expressed the area of the large square in two different ways, now set those two areas equal to each other and then simplify using algebra.
(a+b)×(a+b)=2ab+c2
a2+2ab+b2=2ab+c2
a2+b2=c2
In which step of Betty's proof did she make a mistake? Enter 1 for Step 1, enter 2 for Step 2, enter 3 for Step 3, or enter 4 if Betty did not make a mistake in her proof.
(1 point)
2 answers
Betty made a mistake in Step 1. The area of each right-angled triangle is given by \(\frac{1}{2}bh\), which means the area of such a triangle should be \(\frac{1}{2}ab\), not \(ab\). Therefore, the area of all four triangles should be \(4 \times \frac{1}{2}ab = 2ab\).
So, the correct answer is 1.
